- Show that \(\GF(p^n)\) is the splitting field of \(x^{p^n} - x \in \FF_p[x]\).
Solution
Let \[ f(x)=x^{p^n}-x\in\FF_p[x], \] and let \(L\) be its splitting field over \(\FF_p\).
1. The polynomial \(f\) has exactly \(p^n\) distinct roots in \(L\).
Its derivative is \[ f'(x)=p^n x^{p^n-1}-1=-1 \] in characteristic \(p\). Thus \(\gcd(f,f')=1\), so \(f\) is separable. Since \(\deg f=p^n\) and \(L\) is a splitting field, \(f\) has exactly \(p^n\) distinct roots in \(L\).
2. The root set\[ S=\{a\in L:a^{p^n}=a\} \]is a subfield of \(L\).
Clearly \(0,1\in S\). If \(a,b\in S\), then the Frobenius identity in characteristic \(p\) gives \[ (a-b)^{p^n}=a^{p^n}-b^{p^n}=a-b, \] so \(a-b\in S\), and \[ (ab)^{p^n}=a^{p^n}b^{p^n}=ab, \] so \(ab\in S\). If \(0\ne b\in S\), then \[ (b^{-1})^{p^n}=(b^{p^n})^{-1}=b^{-1}, \] so \(b^{-1}\in S\). Hence \(S\) is a field.
3. The field \(S\) has exactly \(p^n\) elements and contains \(\FF_p\).
By 1, \(S\) is exactly the set of the \(p^n\) distinct roots of \(f\), so \(|S|=p^n\). For \(c\in\FF_p\), one has \(c^p=c\), and iterating Frobenius gives \(c^{p^n}=c\). Thus \(\FF_p\subseteq S\).
4. The splitting field \(L\) equals \(S\).
3. Hence adjoining them produces nothing larger than \(S\), so \(L=S\).
5. Therefore \(L\cong\GF(p^n)\), and \(\GF(p^n)\) is the splitting field of \(x^{p^n}-x\) over \(\FF_p\).
By <1>3–<1>4, \(L\) is a finite field with \(p^n\) elements. Up to unique isomorphism over \(\FF_p\), there is exactly one finite field of order \(p^n\), denoted \(\GF(p^n)\). Thus the splitting field is \(\GF(p^n)\).